Syllabus Edition

First teaching 2023

First exams 2025

|

Inheritance (HL IB Biology)

Exam Questions

4 hours70 questions
1a
Sme Calculator
1 mark

During the process of fertilisation, haploid sperm and egg cells fuse together to form a diploid zygote.

Define the term 'haploid'.

1b
Sme Calculator
2 marks

There are certain advantages to cells being diploid.

List two advantages of cells containing two sets of chromosomes.

1c
Sme Calculator
3 marks

A couple is expecting a child and wondering which sex the baby will be.

The following genetic diagram shows the sex chromosomes present in the gametes of both parents.

Gametes X X
X    
Y    

Calculate the percentage chance of the baby being a girl by completing the genetic diagram. 

Show your working.

1d
Sme Calculator
1 mark

One of the genes carried on the X chromosome codes for a protein called factor VIII.

State the role of this protein in humans.

Did this page help you?

2a1 mark

Define the term genotype.

2b2 marks

In guinea pigs the allele for black hair (B) is dominant to the allele for white hair (b), and the allele for long hair (L) is dominant to the allele for short hair (l).

A double homozygous guinea pig with long, black hair was bred with another double homozygous guinea pig with long, white hair.                          

State the genotypes of the two parent guinea pigs.

2c3 marks

Use a genetic diagram to show the ratio of different phenotypes which could result from the cross discussed in part b).

Did this page help you?

3a
Sme Calculator
3 marks

A group of scientists investigated Mendelian inheritance by carrying out experiments on sweet pea plants. They looked at the inheritance of two traits, flower colour and pollen grain shape.

Flower colour is determined by two alleles; the dominant P for purple and recessive p for red.

Pollen grain shape is determined by two alleles, the dominant L for long and recessive l for round. 

They started by breeding together a double homozygous dominant plant (PPLL) with a double homozygous recessive plant (ppll) to produce a generation of plants that were all double heterozygous (PpLl). 

They then bred the offspring of this first cross together and ended up with 381 offspring in the second generation. 

The outcome of the second round of crosses are as follows:

Phenotype Observed Expected from 9:3:3:1 ratio
Purple flower, long pollen 284  
Purple flower, round pollen 21  
Red flower, long pollen 21  
Red flower, round pollen 55  

Complete the third column of the table above with the expected number of individuals of each phenotype.

3b1 mark

In order to establish whether there was a significant difference between the observed results and the expected results, a statistical test is required.

Identify the statistical test that should be used.

3c1 mark

It was found that there was a significant difference between the expected and observed results from the genetic cross described in part a). 

Suggest a possible reason for the data that was collected in the experiment.

Did this page help you?

4a2 marks

A group of 1000 people were chosen at random and surveyed as part of a population study. The participants were asked about their characteristics. 

One characteristic that was surveyed was the participants' height measurements. 

Sketch a graph in the space below to predict the distribution of height values in the population sample.

Axes for height normal distribution curve SQ

4b1 mark

Name the type of variation shown in the example in part a). 

4c2 marks

Another characteristic that was surveyed was hair colour. 

It was found that most individuals had black, brown, blonde, or red hair, but a small number of individuals had hair colours like pink, blue and green. 

Describe the factors that can cause variation in hair colour.

4d1 mark

Some characteristics are coded for by several genes that work in combination to produce the phenotype. 

State the scientific term used for this type of characteristic. 

Did this page help you?

5a2 marks

Describe autosomal linkage.

5b2 marks

In tomato plants the genes for height and leaf type are autosomally linked.

  • The allele T, for a tall plant, is dominant to the allele t, for a dwarf plant.
  • The allele M, for normal leaves, is dominant to the allele m, for mottled leaves.

A tomato plant is heterozygous with both dominant alleles located on one chromosome and both recessive alleles located on the other. 

Draw the correct notation to represent the genotype of this plant. 

5c3 marks

The plant from part b) was crossed with another plant with the same genotype.

Draw a genetic diagram to predict the genotypes and phenotypes of the offspring produced by this cross. 

5d2 marks

A small number of offspring from the cross described in part c) possess tall, mottled and dwarf, normal phenotypes.

Suggest how it is possible for some plants to have these phenotypes.

Did this page help you?

65 marks

Outline the causes of variation in living organisms.

Did this page help you?

7a
Sme Calculator
1 mark

Huntington's disease is a disease caused by a dominant allele.

State what is meant by the term dominant, in the context of alleles. 

7b
Sme Calculator
3 marks

Gregor Mendel conducted experiments that established the basis of modern genetics. 

(i)
State the type of organism that Mendel used in his studies.

[1]

(ii)
State why the organism you named in part (i) was a good choice for Mendel's experiments.

[2]

7c
Sme Calculator
2 marks

In snapdragon plants, a cross between a red and white flowered plant results in pink flowered offspring due to codominance of alleles.

In the species that Mendel used for his experiments, a cross between a red-flowered plant and a white-flowered plant, did not result in any pink flowered plants in the next generation.

Suggest why. 

Did this page help you?

8
Sme Calculator
1 mark

In a genetic diagram where H denotes the dominant allele responsible for causing Huntington's disease, and h denotes the recessive allele, state:

(i)
the meaning of the genotype Hh
[1]
(ii)
the phenotype that Hh will display

[1]

Did this page help you?

9a
Sme Calculator
2 marks

The diploid number of a species of rodent is 64.

The table below has been incorrectly completed; some of the numbers in the second column are correct, while others are not. 

Cell Type Number of chromosomes
Zygote 64
Sperm cell 64
Muscle cell 32
Fur-producing cell 16

Complete the table below with correct numbers in the second column. 

Cell Type Number of chromosomes
Zygote  
Sperm cell  
Muscle cell  
Fur-producing cell  
9b
Sme Calculator
2 marks

A genetic cross is performed between two heterozygous parents with the genotype Qq.

Complete the Punnett grid for this cross.

punnett-grid---sq

9c
Sme Calculator
2 marks

When writing out genetic crosses by hand, which of the following pairs of letters is the best choice for denoting the dominant and recessive alleles?

Give a reason for your answer. 

  • Cc
  • Oo
  • Hh
  • Vv
9d
Sme Calculator
1 mark

State the name used to describe different alleles of the same gene, that have a combined effect on the phenotype of the organism. 

Did this page help you?

10a
Sme Calculator
2 marks

A pedigree chart is shown below.

MIvO8gH5_e-4a

(i)
State the number of generations shown in the pedigree chart. 

[1]

(ii)
State the numbers of males and females shown in the pedigree chart. 

[1]

10b
Sme Calculator
1 mark

The pedigree chart from part (a) can be adapted with shading as follows:

yh_tdceS_e-4b

Suggest a meaning for the shaded squares and circles. 

10c
Sme Calculator
2 marks

A sperm cell is shown below.mRSrnqqO_e-4c

(i)
Label the diagram with an X to show the position of mitochondria in this cell.

[1]

(ii)
State the purpose of the mitochondria in a sperm cell.

[1]

10d
Sme Calculator
2 marks

Define the term sex linkage. 

Did this page help you?

11a
Sme Calculator
4 marks

State the differences between the terms phenotype and genotype. 

Give one example of each. 

11b
Sme Calculator
3 marks

8000 offspring were produced from a cross between two heterozygous parents, Hh and Hh. 

Calculate how many of these offspring would have the genotype hh. Show your working. 

Did this page help you?

1a2 marks

Hair colour and eye colour are traits controlled by genes. Many people with blonde hair also have blue eyes.

Suggest a reason for this observation.

1b3 marks

In the common pea plant (Pisum sativum) the allele P, for purple flowers, is dominant to the allele p, for white flowers. The allele I, for inflated seed pods, is dominant to the allele i, for constricted seed pods.

Two pea plants, both heterozygous for both characteristics are crossed.

Complete a Punnett square to show the possible genotypes of the offspring. 

1c2 marks

The genetic cross described above produced 112 seeds.

Calculate the expected number of offspring with white flowers and constricted seed pods.

Did this page help you?

2a2 marks

The genes for body colour and antenna shape in fruit flies (Drosophila melanogaster) are close together on the same chromosome. These genes are therefore said to be linked.

Allele E for a striped body is dominant over allele e for an ebony body.

Allele A codes for the dominant, normal antennae, whereas allele a codes for an aristopedia antennae. Aristopedia antennae resemble a Drosophila leg rather than an antenna.

A male that is heterozygous for striped body and normal antennae is crossed with a female that has an ebony body and aristopedia antennae.

State the possible allele combinations in the gametes of these flies. Use the correct notation in your answer.

2b2 marks

State the possible phenotypes of the offspring of the cross described in part a), along with the predicted phenotype ratios. 

2c2 marks

When carried out, the cross described in part a) produced offspring with the following phenotypes:

Phenotype No. of individuals
Striped body and normal antennae 125
Ebony body and aristopedia antennae 122
Striped body and aristopedia antennae 12
Ebony body and normal antennae 9

Explain the observed phenotypes.

Did this page help you?

3a2 marks

In fruit flies (Drosophila melanogaster) wing length and body colour are each controlled by a single gene with two alleles.

Allele L for long wings is dominant over allele l for vestigial wings, while allele G for grey body colour is dominant over allele g for ebony body colour.

Two homozygous fruit flies were crossed; one had a grey body and long wings while the other had an ebony body and vestigial wings.

State, with a reason, the number of offspring that would be expected to display a grey body colour and vestigial wings if 350 offspring were produced from this cross.

3b1 mark

Two fruit flies from the cross in part a) were crossed and 3 200 offspring were produced.

Calculate the expected number of offspring that would display the following phenotypes, assuming that the genes for body colour and wing length are not linked.

Phenotype Expected number of offspring
Grey body, long wings  
Grey body, vestigial wings  
Ebony body, long wings  
Ebony body, vestigial wings  
3c4 marks

The results for the cross in part a) were different to the expected values in part b). Scientists decide to perform a chi-squared test to determine whether or not this difference is significant.

Calculate χ2 by completing the table below. Note that the formula has been provided.

Phenotype of offspring Observed (O) Expected (E) left parenthesis O minus E right parenthesis left parenthesis O minus E right parenthesis squared left parenthesis O minus E right parenthesis squared over E
Grey body, long wings 1 650        
Grey body, vestigial wings 600        
Ebony body, long wings 690        
Ebony body, vestigial wings 260        
        capital sigma left parenthesis O minus E right parenthesis squared over E=  

3d3 marks

The table below shows critical values for the χ2 test.

q4d_10-2_inheritance_medium_ib_hl_biology_sq

Use the χ2 value from part c) and the critical values table to state a conclusion for this cross.

Did this page help you?

4a4 marks

Contrast continuous and discontinuous variation.

4b4 marks

Explain the use of a chi-squared test.

4c3 marks

Describe sex linkage.

Did this page help you?

5a2 marks

Human red blood cells can be categorised into different blood groups based on the structure of a surface glycoprotein (antigen). The ABO blood groups are controlled by a single gene with multiple alleles (A, B, O). The table below shows all the genotypes for all the possible blood groups.

Phenotype Genotype
Blood Group A IAIA    IAi
Blood Group B IBIB    IBi
Blood Group AB IAIB
Blood Group O ii


A child has blood group AB and their father has blood group A.
 

Identify the possible phenotypes of the mother.

5b1 mark

Suggest which pattern of inheritance is exhibited in the AB blood group.

5c3 marks

A man with normal colour vision wishes to start a family with his partner who is a carrier of the allele for colour-blindness. They wanted to work out the probability of having a child with colour-blindness.

Using the following symbols: 

XB = an X chromosome carrying the normal allele for colour vision
Xb = an X chromosome carrying the allele for colour blindness

(i)
Identify all the possible genotypes for female and male offspring.

(ii)
Predict the probability of having a child with colour-blindness.

Did this page help you?

6a2 marks

Hypophosphatemia is a sex-linked inherited condition which results in abnormally low levels of phosphate in the blood which can cause the disease rickets. It is caused by a dominant allele.

The diagram below shows the inheritance of hypophosphatemia in one family.

biology-3-3-q2a-ib-hl-sq

State the evidence that suggests that hypophosphatemia is a sex-linked, dominant inherited disease.

6b2 marks

Using the following symbols,

         XH = an X chromosome carrying the allele for hypophosphatemia
         Xh = an X chromosome carrying the normal allele
         Y = a Y chromosome 

Identify all the possible genotypes of each of the following persons from the diagram in part (a):

 1  :

 4  :

 5  :

13 :

6c2 marks

Person 20, from the diagram in part (a), is pregnant for the fourth time. As the family has a history of hypophosphatemia, a test was carried out to discover the sex of the embryo. 

Describe what possible observations of the chromosomes would be expected when determining the sex of an embryo.

6d4 marks

State the probability that the child Person 20 is pregnant with will be a male with hypophosphatemia.

Explain your answer by drawing a genetic diagram, using the following symbols:

 XH = an X chromosome carrying the allele for hypophosphatemia

Xh = an X chromosome carrying the normal allele

Y = a Y chromosome

Did this page help you?

7a2 marks

A horticulturist investigated the inheritance of flower colour in Camellia japonica, a widely cultivated ornamental plant commonly known as Japanese camellia.The horticulturist crossed a homozygous parent with red flowers and a homozygous parent with white flowers. All of the F1 generation had the same colour flowers. Using the following symbols: 

CR  = Red flowers
CW = White flowers 

Sketch a genetic diagram / Punnett square to deduce all the genotypes in this cross

7b2 marks

Each of the F1 generation plants had flowers that were patterned red and white. The horticulturist undertook a self-cross with these F1 hybrids.

(i)
State all the possible phenotypes of the F2 hybrids

(ii)
Deduce the probability of obtaining a white flower.
7c3 marks

Describe, with a reason, what pattern of inheritance is exhibited in the horticulturist’s experiment.

Did this page help you?

8a2 marks

FG syndrome is a recessive disorder that can cause a characteristic facial appearance, developmental delays and hyperactivity. FG syndrome is a rare X-linked genetic disorder that occurs almost exclusively in males, it is caused by a mutation in the MED12 gene on the X chromosome.

Suggest why FG syndrome occurs almost exclusively in males.

8b3 marks

The diagram below shows the familial inheritance of an X-linked recessive disease.

biology-q4b-3-3-sq-ib-hl

Describe the patterns of inheritance that hold true for X-linked conditions.

8c1 mark

Haemophilia is due to a sex-linked recessive gene Xh  whereas the normal gene is XH. A haemophiliac man and a woman, who does not have haemophilia, have two children. Their first child is male and has haemophilia. 

Deduce what this tells us about the mother.

8d2 marks

Their second child is female. 

Deduce, with a reason, the probability that their daughter will also have haemophilia.

Did this page help you?

17 marks

Outline the inheritance of colour-blindness.

Did this page help you?

2a
Sme Calculator
1 mark

Human eggs and sperm cells are very different in size to each other.

Explain why, despite this size difference, both contribute equally to the genetic composition of a zygote. 

2b
Sme Calculator
3 marks

Explain the concept of, and the importance of, segregation of alleles. 

Did this page help you?

3a
Sme Calculator
2 marks

Explain why knowledge of blood groups is of critical importance when planning a blood transfusion. 

3b
Sme Calculator
3 marks

A man of blood group AB and a woman of blood group B have four children together. 

  • One child is blood group AB
  • One child is blood group A
  • Two children are blood group B

Assuming that the genotypes of the four children are representative of the expected genotype ratios, deduce the mother's and father's genotypes under the A, B, O blood grouping system.

3c
Sme Calculator
2 marks

Using your knowledge of the A, B, O blood grouping system, suggest why people of blood group O are sought-after as blood donors. 

3d
Sme Calculator
2 marks

As stated in part (c), group O blood is highly valued for transfusions into other patients. 

Explain the main disadvantage of a person having group O blood. 

Did this page help you?

4a
Sme Calculator
4 marks

Chickens can produce pigmentation in their feathers to make them white, black or speckled, as shown in the diagram below.

2Sv3tp0I_h-3a

Homozygous white-feathered chickens can be crossed with homozygous black-feathered chickens to produce speckled offspring. This occurs via codominance. 

Construct a Punnett grid to show the results of two of the speckled offspring being crossed. Use your Punnett grid to deduce the ratios of the various phenotypes that would come out of the cross. 

4b
Sme Calculator
2 marks

Explain why the traits shown in part (a) are referred to as codominant.

4c
Sme Calculator
2 marks

The genetic disease sickle cell anaemia is caused by a faulty allele of the beta-globin gene, needed for the production of functional haemoglobin in red blood cells. 

The faulty allele causes red blood cells to adopt a sickle shape when oxygen availability is low, as opposed to the conventional biconcave disc structure, as shown below. Sickle cells cause considerable suffering including severe cramping pains in fingers and toes, and general fatigue.

-3CRTVfS_h-3c

The following table gives information about the condition in its various forms.

Genotype Disease manifestation
Homozygous dominant
No disease
All red blood cells are biconcave discs
Heterozygous
Mild symptoms at worst
Mixture of biconcave discs and sickle-shaped cells at low oxygen levels
Homozygous recessive
Severe disease
All red blood cells sickle-shaped at low oxygen levels, considerable suffering

Use the information above to explain why the condition is regarded as codominant. 

Did this page help you?

5a
Sme Calculator
2 marks

Explain how mutations can have a positive effect on a species. 

5b
Sme Calculator
3 marks

Galactosaemia is a condition that prevents sufferers from fully metabolising the sugar galactose.

The pedigree chart below shows part of a family in which galactosaemia is an inherited condition.

Jw3~6u8K_h-4b

Explain how this pedigree chart indicates whether galactosaemia is recessive, sex-linked or both. 

Did this page help you?

6a
Sme Calculator
7 marks


(i)
A certain species of flower grows with either red or white petals. The allele for red flowers, R, is dominant to the allele for white flowers, r. 

Construct a genetic diagram to predict the outcome of crossing pure-bred red flowers with pure-bred white flowers.

State the genotype and phenotype ratios that would be expected in the F1 generation. 

[3]

(ii)

Plants from the F1 generation were crossed. 

Construct a second genetic diagram to predict the outcomes.

State the genotype and phenotype ratios that would be expected in the F2 generation. 

[4]

6b
Sme Calculator
4 marks

Huntington's disease is a genetic condition that affects the brain progressively. Problems with coordination worsen over time and can ultimately cause death by, for example, an inability to swallow or by injuries associated with falling. 

(i)

Huntington's disease is described as an autosomal dominant disorder.

Explain the term autosomal dominant.

[2]

(ii)

The mutation in the Huntington's disease allele contains a higher-than-normal number of repeats of a nucleotide sequence coding for a protein called huntingtin.

Suggest the consequence to the protein huntingtin of the extra nucleotide repeats. 

[2]

Did this page help you?

7a3 marks

Scientists found that the genes controlling body colour and wing length in fruit flies, Drosophila melanogaster, are inherited on different autosomal chromosomes. Fruit flies are either black or grey and have either long or short wings. 

A homozygous fruit fly with a black body and long wings was crossed with a homozygous grey fruit fly with short wings. All of the offspring produced had black bodies with long wings. 

Use a genetic diagram to show how these offspring were produced. Use B/b and L/l to represent the alleles. 

7b3 marks

The offspring from part a) were crossed with grey bodied fruit flies with short wings.

Use a genetic diagram to deduce the expected phenotype ratios from this cross.

7c1 mark

The observed phenotypes from the cross in part b) are shown in the table below:

Phenotype Observed Number Expected Number
Black body and long wings 83  
Grey body and long wings 78  
Black body and short wings 85  
Grey body and short wings 74  

Complete the table to show the expected number of offspring for each phenotype.

7d5 marks

The chi-squared test can be used to determine whether or not the different between observed and expected values is signficant.

The formula below can be used to determine the value of chi-squared.

chi-squared-sq

(i)

Calculate the value of chi-squared (χ2), using the above formula.

[2]

(ii)

The critical value for this data is 7.82.

Use this information and your answer to part (i) to deduce a conclusion from this chi-squared test.

[3]

Did this page help you?

8a2 marks

When investigating variation scientists often study sets of twins, both identical and non-identical. 

Suggest the advantage of studying twins when investigating variation.

8b3 marks

A scientific study carried out in 2013 investigated whether academic achievement was influenced primarily by genetics or the environment. 

11 117 pairs of identical twins participated in the study, which found that around 55% of the outcomes in core GCSE subjects (English, maths, and science) were explained by genetic influence, 25% by shared environmental influences, such as parental support, and the remaining 20% by environmental influences that were not shared between the two twins, such as teacher quality and class grouping. 

Using this information, evaluate the statement "intelligence is determined by genetics". 

8c2 marks

Twin studies tend to be run using two main assumptions:

  • That the identical twins in the study share 100% of their DNA
  • That all twins are raised in exactly equal environments at home

Discuss whether it is correct to make these assumptions about twin studies.

Did this page help you?

9a3 marks

The diagram below shows a pair of homologous chromosomes from the fruit fly, Drosophila melanogaster. 

Linked genes on homologous chromosomes SQ

The white regions are the loci of seven genes involved in different phenotypic traits. 

The letters A-F and a-f represent the alleles present at each locus. 

Discuss the relative chances of this fly's gametes containing these combinations of alleles:

  • A and F
  • A and f
  • c and d
  • c and D
9b3 marks

Drosophila melanogaster is a useful organism to use in studies on inheritance patterns. Female fruit flies can lay up to 400 eggs, which can develop into adults in 7-14 days. They have simple nutrient requirements.

Explain why Drosophila melanogaster is a useful organism in studies of inheritance patterns.

Did this page help you?

104 marks

Mendel studied many characteristics in pea plants. Two such characteristics are seed colour and flower colour.

  • Seed colour can either be yellow, Y, or green, y. 
  • Flower colour can either be purple, P, or white, p. 

Outline a method they could be used to determine whether the two genes are linked. 

Did this page help you?

11
Sme Calculator
7 marks

Pollen from a pure-breeding tomato plant with white flowers and yellow fruit was transferred to the stigmas of a pure-breeding plant with yellow flowers and red fruit. All the F1 generation had yellow flowers and red fruit. 

Pollen from the F1 generation was transferred to pure-breeding plants with white flowers and yellow fruit. The ratio of phenotypes expected among the offspring of a dihybrid test cross such as this is 1:1:1:1.

Seeds from the plants were collected and grown, giving plants with the following phenotypes:

  • Yellow flowers and red fruit: 59
  • Yellow flowers and yellow fruit: 56
  • White flowers and red fruit: 40
  • White flowers and yellow fruit: 45

A chi-squared (χ2) test can be carried out to check whether the numbers of each phenotype match the expected 1:1:1:1 ratio. 

The chi-squared (χ2) equation and critical values are as follows:

chi-squared-sq

Degrees of freedom Probability (p)
0.10 0.05 0.02 0.01 0.001
1 2.71 3.84 5.41 6.64 10.83
2 4.61 5.99 7.82 9.21 13.82
3 6.25 7.82 9.84 11.35 16.27
4 7.78 9.49 11.67 13.28 18.47

Use this information to deduce a conclusion about the difference between the expected and actual results in the tomato breeding experiment.

You must show your working in your answer.

Did this page help you?